$N=\langle g^2\mid g \in G \rangle$ is normal in $G$

Let $G$ be a group and let $N=\langle g^2\mid g \in G \rangle$.

Show that $N \lhd G$ and $G/N$ is abelian.

Well, I've come up with a simple approach:

If $n=x^2 \in N$ for some $x \in G$ then $n^g = gng^{-1}=(gxg^{-1})^2 \in N$ and from here it is simple to prove for a general $n \in N$.

Now I'm trying another solution: I define a function $f:G \rightarrow \{0,1\}$ so that $f(g) = 1$ if $g \in N$ and $f(g) = 0$ otherwise. If this is an homomorphism than $N=Kerf$ and we are done. I'm having trouble however to prove that if $x,y \notin N$ than $f(xy)=1$ (meaning that $xy \in N$). If this is true, I'd be happy for some clues.

Thanks.

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Why is $N$ even a subgroup? – ronno Aug 30 '12 at 19:52
@ronno it's a subgroup by definition. – Dustan Levenstein Aug 30 '12 at 19:53
@ronno: $N$ is not $\{g^2:g\in G\}$; it’s the subgroup of $G$ generated by that set. – Brian M. Scott Aug 30 '12 at 19:54
Working with $f:G\to\{0,1\}$ will not work, as $G:N$ need not be 2 as even in the abelian case the product of non-squares need not be square. For example, with the multiplicative group of non-zero rationals $G=\mathbb Q^\times$ the index is infinite and $G/N$ is the sum of countably many copies of $\mathbb Z_2$ (one for each prime number and $-1$). – Hagen von Eitzen Sep 1 '12 at 21:11

2 Answers

The proposition that when you multiply two elements not in $N$ together, you get an element in $N$ is not true in general. Consider the Klein Four group, $G = \langle a,b \mid a^2, b^2, (ab)^2\rangle$, and let $N$ be defined on $G$. Then $a, b \notin N$, and $ab \notin N$.

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Roy proved that $N$ is normal. By definition of <> the set $N$ is a subgroup. Then every element of $G/N$ has order dividing 2 so it is abelian.

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@DustanLevenstein Since the quotient $G/N$ is abelian, with every element of order 2, then it is isomorphic to a direct sum of copies of $Z_2$; there are many choices of a homomorphism from this direct sum to $Z_2$. – i. m. soloveichik Aug 30 '12 at 23:01